If the circle `x^2+y^2+2gx+2fy+c=0` is touched by `y=x` at `P` such that `O P=6sqrt(2),` then the value of `c` is (a)36 (b) 144 (c) 72 (d) none of these

Options

A. 36
B. 144
C. 72
D. none of these

Answer

3

Explanation

The equation of the line `y=x` in distance form is
`(x)/(cos theta)=(y)/(sin theta)=r`, where ` theta =(pi)/(4)`
For point P`, r=6 sqrt(2)`. Therefore, the coordinates of P are given by
`(x)/(cos (pi//4))=(y)/(sin (pi//4))=6 sqrt(2)` or `x=6,y=6`
Since `y=x` touches the circle, the equation `2x^(2)+2x(g+f)+c=0` has the equal roots . Therefore,
`4(g+f)^(2)=8c`
or `(g+f)^(2)=2c` (2)
From (1), we get
`[12(g+f)^(2)=[-(c+72)]^(2)`
or `144(g+f)^(2)=(c+72)`
or`144(2c)=(c+72)^(2)`
or `(c-72)^(2)=0` or `c=72`